Question

Express $y=\textrm{tan}\left(\frac{\sqrt{1-x^{2}}}{x}\right)$ in terms of Cos inverse x

Answer 1

The expression in terms of $\cos^{-1} x$ is $\tan(\tan(\arccos x))$.

Answer 2

Let $\theta = \arccos x$. Then $x = \cos \theta$ and $\sqrt{1-x^2} = \sin \theta$ for $\theta \in [0, \pi]$. Substituting these into the given equation $y=\tan\left(\frac{\sqrt{1-x^{2}}}{x}\right)$ yields $y=\tan\left(\frac{\sin \theta}{\cos \theta}\right) = \tan(\tan \theta)$. Replacing $\theta$ with $\arccos x$ gives the final answer.